Question

The graph of a cubic polynomial function y = f(x) is shown. It is known that one of the zeros is...

The graph of a cubic polynomial function
y = f(x)
is shown. It is known that one of the zeros is
1 + i.
Write an equation for f.
0 0
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Answer #1

you haven't shown the graph, but I suppose it would be showing the ...

one and ONLY x intercept! Take that value, and turn it into a factor.

For instance, if the graph intersected at -1, then your factor would be (x- -1) = (x+1).

The given imaginary zeros, in your case 1 + i,ALWAYS come in conjugate PAIRS.

That mean that you have another: 1 -i cool, huh

turn each of those into a factor:

[ x - (1 - i)] = [ x - 1 + i ] and...

[ x - (1 + i)] = [ x - 1 - i ]

regrouping these allows you to get...

[ (x-1) + i ] & [ (x-1) - i ] (**)

you might recognize this as the pattern (a-b) &(a+b) which you can multiply and get... a2 - b2

This coming back...??

so take the terms at (**) and multiply them to get...

[ (x-1)2 - i2 ] which is...

x2 - 2x + 1 + 1 = x2 - 2x + 2 cool. so the two imaginary roots gave us:

(x2 - 2x + 2)

multiply the above by the root you see in your graph (that you didn't show)... for example:

(x - a) (x2 - 2x + 2)

When you are done, you will have a cubic equation...

f(x) = x3 + etc. etc....

that's about it! Let me know how it goes!

answered by: I need some explaining
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